invariant subspaces for selfadjoint *algebras of operators
In this entry we provide few results concerning invariant subspaces of *algebras of bounded operators^{} on Hilbert spaces^{}.
Let $H$ be a Hilbert space and $B(H)$ its algebra of bounded operators. Recall that, given an operator $T\in B(H)$, a subspace^{} $V\subseteq H$ is said to be invariant^{} for $T$ if $Tx\in V$ whenever $x\in V$.
Similarly, given a subalgebra $\mathcal{A}\subseteq B(H)$, we will say that a subspace $V\subseteq H$ is invariant for $\mathcal{A}$ if $Tx\in V$ whenever $T\in \mathcal{A}$ and $x\in V$, i.e. if $V$ is invariant for all operators in $\mathcal{A}$.
Invariant subspaces for a single operator
Proposition^{} 1  Let $T\mathrm{\in}B\mathit{}\mathrm{(}H\mathrm{)}$. If a subspace $V\mathrm{\subset}H$ is invariant for $T$, then so is its closure^{} $\overline{V}$.
Proof: Let $x\in \overline{V}$. There is a sequence $\{{x}_{n}\}$ in $V$ such that ${x}_{n}\to x$. Hence, $T{x}_{n}\to Tx$. Since $V$ is invariant for $T$, all $T{x}_{n}$ belong to $V$. Thus, their limit $Tx$ must be in $\overline{V}$. We conclude that $\overline{V}$ is also invariant for $T$. $\mathrm{\square}$
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Proposition 2  Let $T\mathrm{\in}B\mathit{}\mathrm{(}H\mathrm{)}$. If a subspace $V\mathrm{\subset}H$ is invariant for $T$, then its orthogonal complement^{} ${V}^{\mathrm{\u27c2}}$ is invariant for ${T}^{\mathrm{*}}$.
Proof: Let $y\in {V}^{\u27c2}$. For all $x\in H$ we have that $\u27e8x,{T}^{*}y\u27e9=\u27e8Tx,y\u27e9=0$, where the last equality comes from the fact that $Tx\in V$, since $V$ is invariant for $T$. Therefore ${T}^{*}y$ must belong to ${V}^{\u27c2}$, from which we conclude that ${V}^{\u27c2}$ is invariant for ${T}^{*}$. $\mathrm{\square}$
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Proposition 3  Let $T\mathrm{\in}B\mathit{}\mathrm{(}H\mathrm{)}$, $V\mathrm{\subset}H$ a closed subspace and $P\mathrm{\in}B\mathit{}\mathrm{(}H\mathrm{)}$ the orthogonal projection onto $V$. The following are statements are equivalent^{}:

1.
$V$ is invariant for $T$.

2.
${V}^{\mathrm{\u27c2}}$ is invariant for ${T}^{\mathrm{*}}$.

3.
$T\mathit{}P\mathrm{=}P\mathit{}T\mathit{}P$.
Proof: $(1)\u27f9(2)$ This part follows directly from Proposition 2.
$(2)\u27f9(1)$ From Proposition 2 it follows that ${({V}^{\u27c2})}^{\u27c2}$ is invariant for ${({T}^{*})}^{*}=T$. Since $V$ is closed, $V=\overline{V}={({V}^{\u27c2})}^{\u27c2}$. We conclude that $V$ is invariant for $T$.
$(1)\u27f9(3)$ Let $x\in H$. From the orthogonal decomposition theorem we know that $H=V\oplus {V}^{\u27c2}$, hence $x=y+z$, where $y\in V$ and $z\in {V}^{\u27c2}$. We now see that $TPx=Ty$ and $PTPx=PTy=Ty$, where the last equality comes from the fact that $Ty\in V$. Hence, $TP=PTP$.
$(3)\u27f9(1)$ Let $x\in V$. We have that $Tx=TPx=PTPx$. Since $PTPx$ is obviously on the image of $P$, it follows that $Tx\in V$, i.e. $V$ is invariant for $T$. $\mathrm{\square}$
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Proposition 4  Let $T\mathrm{\in}B\mathit{}\mathrm{(}H\mathrm{)}$, $V\mathrm{\subset}H$ a closed subspace and $P\mathrm{\in}B\mathit{}\mathrm{(}H\mathrm{)}$ the orhtogonal projection onto $V$. The subspaces $V$ and ${V}^{\mathrm{\u27c2}}$ are both invariant for $T$ if and only if $T\mathit{}P\mathrm{=}P\mathit{}T$.
Proof: $(\u27f9)$ From Proposition 3 it follows that $V$ is invariant for both $T$ and ${T}^{*}$. Then, again from Proposition 3, we see that $PT={({T}^{*}P)}^{*}={(P{T}^{*}P)}^{*}=PTP=TP$.
$(\u27f8)$ Suppose $TP=PT$. Then $PTP=TPP=TP$, and from Proposition 3 we see that $V$ is invariant for $T$.
We also have that $P{T}^{*}={T}^{*}P$, and we can conclude in the same way that $V$ is invariant for ${T}^{*}$. From Proposition 3 it follows that ${V}^{\u27c2}$ is also invariant for $T$. $\mathrm{\square}$
Invariant subspaces for *algebras of operators
We shall now generalize some of the above results to the case of selfadjoint^{} subalgebras of $B(H)$.
Proposition 5  Let $\mathrm{A}$ be a *subalgebra of $B\mathit{}\mathrm{(}H\mathrm{)}$ and $V$ a subspace of $H$. If a subspace $V$ is invariant for $\mathrm{A}$, then so are its closure $\overline{V}$ and its orthogonal complement ${V}^{\mathrm{\u27c2}}$.
Proof: From Proposition 1 it follows that $\overline{V}$ is invariant for all operators in $\mathcal{A}$, which means that $V$ is invariant for $\mathcal{A}$.
Also, from Proposition 2 it follows that ${V}^{\u27c2}$ is invariant for the adjoint^{} of each operator in $\mathcal{A}$. Since $\mathcal{A}$ is selfadjoint, it follows that ${V}^{\u27c2}$ is invariant for $\mathcal{A}$. $\mathrm{\square}$
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Theorem  Let $\mathrm{A}$ be a *subalgebra of $B\mathit{}\mathrm{(}H\mathrm{)}$, $V\mathrm{\subset}H$ a closed subspace and $P$ the orthogonal projection onto $V$. The following are equivalent:

1.
$V$ is invariant for $\mathrm{A}$.

2.
${V}^{\mathrm{\u27c2}}$ is invariant for $\mathrm{A}$.

3.
$P\mathrm{\in}{\mathrm{A}}^{\mathrm{\prime}}$, i.e. $P$ belongs to the commutant of $\mathrm{A}$.
Proof: $(1)\u27fa(2)$ This equivalence follows directly from Proposition 5 and the fact that $V$ is closed.
$(1)\u27f9(3)$ Suppose $V$ is invariant for $\mathcal{A}$. We have already proved that ${V}^{\u27c2}$ is also invariant for $\mathcal{A}$. Thus, from Proposition 4 it follows that $P$ commutes with all operators in $\mathcal{A}$, i.e. $P\in {\mathcal{A}}^{\prime}$.
$(3)\u27f9(1)$ Suppose $P\in {\mathcal{A}}^{\prime}$. Then $P$ commutes with all operators in $\mathcal{A}$. From Proposition 4 it follows that $V$ is invariant for each operator in $\mathcal{A}$, i.e. $V$ is invariant for $\mathcal{A}$. $\mathrm{\square}$
Title  invariant subspaces for selfadjoint *algebras of operators 

Canonical name  InvariantSubspacesForSelfadjointalgebrasOfOperators 
Date of creation  20130322 18:40:23 
Last modified on  20130322 18:40:23 
Owner  asteroid (17536) 
Last modified by  asteroid (17536) 
Numerical id  9 
Author  asteroid (17536) 
Entry type  Feature 
Classification  msc 46K05 
Classification  msc 46H35 